[网络流24题]圆桌问题

Posted 2021-02-07 birchtree

[luogu 3254] 圆桌问题

传送门

二分图多重匹配,将单位看作左部节点，餐桌看作右部节点

限制为每个左部节点最多连r[i]条边，每个右部节点最多连c[i]条边

最大流即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#define maxn 1005
#define maxm 100005
#define INF 0x7fffffff
using namespace std;
int n,m;
int r[maxn];
int c[maxn];

struct edge{
    int from;
    int to;
    int next;
}E[maxm];
int head[maxn];
int sz=1;
int vol[maxm];
int flow[maxm];
void add_edge(int u,int v,int w){
    sz++;
    E[sz].from=u;
    E[sz].to=v;
    E[sz].next=head[u];
    head[u]=sz;
    vol[sz]=w;
    flow[sz]=w;
}

queue<int>q;
int deep[maxn];
int bfs(int s,int t){
    while(!q.empty()) q.pop();
    memset(deep,0,sizeof(deep));
    deep[s]=1;
    q.push(s);
    while(!q.empty()){
        int x=q.front();
        q.pop();
        for(int i=head[x];i;i=E[i].next){
            int y=E[i].to;
            if(flow[i]&&!deep[y]){
                deep[y]=deep[x]+1;
                q.push(y);
                if(y==t) return 1;
            }
        }
    } 
    return 0;
}

int dfs(int x,int t,int minf){
    if(x==t) return minf;
    int rest=minf,k;
    for(int i=head[x];i;i=E[i].next){
        int y=E[i].to;
        if(flow[i]&&deep[y]==deep[x]+1){
            k=dfs(y,t,min(rest,flow[i]));
            if(k==0) deep[y]=0;
            flow[i]-=k;
            flow[i^1]+=k;
            rest-=k;
            if(!rest) break;
        }
    }
    return minf-rest;
}

int dinic(int s,int t){
    int maxflow,nowflow;
    maxflow=0;
    while(bfs(s,t)){
        while(nowflow=dfs(s,t,INF)) maxflow+=nowflow;
    }
    return maxflow;
}

vector<int>ans;
void print_ans(){
    for(int i=1;i<=m;i++){
        ans.clear();
        for(int j=head[i];j;j=E[j].next){
            if(flow[j]==vol[j]-1&&E[j].to>=m&&E[j].to<=n+m){
//              printf("%d ",E[j].to-m);
                ans.push_back(E[j].to-m);
            } 
        }
        sort(ans.begin(),ans.end());
        for(int i=0;i<ans.size();i++){
            printf("%d ",ans[i]); 
        } 
        printf("
");
    }
}
int main(){
    int tot=0;
    scanf("%d %d",&m,&n);
    for(int i=1;i<=m;i++) scanf("%d",&r[i]);
    for(int i=1;i<=n;i++) scanf("%d",&c[i]);
    for(int i=1;i<=m;i++){
        for(int j=1;j<=n;j++){
            add_edge(i,m+j,1);
            add_edge(m+j,i,0);
        }
    }
    for(int i=1;i<=m;i++){
        tot+=r[i];
        add_edge(0,i,r[i]);
        add_edge(i,0,0);
    }
    for(int i=1;i<=n;i++){
        add_edge(m+i,m+n+1,c[i]);
        add_edge(m+n+1,m+i,0);
    }
    if(dinic(0,m+n+1)==tot){
        printf("1
");
        print_ans();
    }else{
        printf("0
");
    }
}